If A is an acute angle and sec A = 17/8, find all other trigonometric ratios of angle A (using
trigonometric identities)
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Given, secA=17/8 and A is an acute angle.
So, in ΔABC we have ∠B=90
0
,
And, AC=17 and AB=8,
By Pythagoras theorem,
BC=
(AC
2
−AB
2
)
=
(17
2
−8
2
)
=
(289−64)
=
225
=15.
Now,
sinA=BC/AC=15/17
cosA=1/secA=8/17
tanA=Bc/AB=15/8
cotA=1/tanA=8/15
cosecA = 1/sinA=17/15.
So, in ΔABC we have ∠B=90
0
,
And, AC=17 and AB=8,
By Pythagoras theorem,
BC=
(AC
2
−AB
2
)
=
(17
2
−8
2
)
=
(289−64)
=
225
=15.
Now,
sinA=BC/AC=15/17
cosA=1/secA=8/17
tanA=Bc/AB=15/8
cotA=1/tanA=8/15
cosecA = 1/sinA=17/15.
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